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a) (3x+y)(9x2-3xy+y2) = 27x3+y3
b)(2x-5)(4x2+10x+25) = 8x3-125
| a)(3x+y)(9x2 -3xy+y2 )=27x3 +y3 |
| b)(2x-5)(4x2 +10x+52 )=8x3 -125 |
| nho | k |
| minh | nhe |
a) Ta có: 27\(x^3\)+ y\(^3\) = (3x)\(^3\) + y\(^3\)= (3x + y)[(3x)\(^2\) – 3x . y + y\(^2\)] = (3x + y)(9x\(^2\) – 3xy + y\(^2\))
Nên: (3x + y) (9x\(^2\) – 3xy + y\(^2\)) = 27x\(^3\) + y\(^3\)
b) Ta có: 8x\(^3\) – 125 = (2x)\(^3\) – 53= (2x – 5)[(2x)\(^2\) + 2x . 5 + 5\(^2\)]
= (2x – 5)(4x\(^2\) + 10x + 25)
Nên:(2x – 5)(4x\(^2\) + 10x + 25)= 8x\(^3\) – 125
Trả lời:
a) Ta có:
27x3 + y3 = (3x)3 + y3= (3x + y)[(3x)2 – 3x . y + y2] = (3x + y)(9x2 – 3xy + y2)
Nên: (3x + y) (9x2 – 3xy + y2 ) = 27x3 + y3
b) Ta có:8x3 - 125 = (2x)3 - 53= (2x - 5)[(2x)2 + 2x . 5 + 52]
= (2x - 5)(4x2 + 10x + 25)
Nên: (2x - 5)(4x2+ 10x +25 ) = 8x3 - 125
a) (3x +y)( 9x2 - 3xy + y2)
b) (2x - 5)( 4x2 + 10x + 25)
(đề bài này hay đó)
\(a,\left(3x+y\right)\left(9x^2-3xy+y^2\right)=27x^3+y^3\)
\(b,\left(2x-5\right)\left(4x^2+10x+25\right)=8x^3-125\)
Bài 1 : Khai triển .
a, \(\left(2x-y\right)^3=8x^3-12x^2y+6xy^2-y^3\)
b, \(27x^3+1=\left(3x+1\right)\left(9x^2-3x+1\right)\)
c, \(\left(2x\right)^3-125=\left(2x-5\right)\left(4x^2+10x+25\right)\)
Bài 2 :
a, \(x^2+10x+25=\left(x+5\right)^2=\left(x+5\right)\left(x+5\right)\)
b, \(16x^2-\left(x^2+4\right)^2=\left(4x-x^2+4\right)\left(4x+x^2+4\right)\)
c, \(x^2+3x^2+3x+1=4x^2+3x+1\)
\(=\left(2x\right)^2+\frac{2x.3}{2}+\frac{9}{4}-\frac{5}{4}=\left(2x+\frac{3}{2}\right)^2-\left(\sqrt{\frac{5}{4}}\right)^2\)
\(=\left(2x+\frac{3}{2}-\sqrt{\frac{5}{4}}\right)\left(2x+\frac{3}{2}+\sqrt{\frac{5}{4}}\right)\)
bai2 :cmr
a, a^3+b^3=(a+b)^3-3ab.(a+b)
VP= \(\left(a+b\right)^3-3ab\left(a+b\right)\)
=\(a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2=a^3+b^3\)
=VT
b.a^3-b^3=(a-b)^3+3ab,(a-b)
\(VP=\left(a-b\right)^3+3ab\left(a-b\right)\)
=\(a^3-3a^2b+ab^2.3-b^3+3a^2b-3ab^2=a^3-b^3\)
=VT
=> ĐPCM
bài 1.
a) = 8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3-(8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3)
= 8x3+4x2y+2xy2-4x2y-2xy2-y3 - 8x3+4x2y-2xy2-4x2y+2xy2-y3
=-8x2y-6y3
b) = 27x3-18x2y+12xy2+18x2y-12xy2+8y3-27x3
=8y
a: \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)=27x^3+y^3\)
b: \(\left(2x-5\right)\left(4x^2+10x+25\right)=8x^3-125\)